Quadratic extensions of the 2-adic numbers (${\mathbb{Q}}_2(\sqrt{d}),d=-1,5,-5,2,-2,10,-10$)

The three generators $-1,2,5$ of the square-class group of the 2-adic numbers give these seven distinct extensions. In characteristic different from two, completing the square shows every quadratic extension is of this form, and comparing coefficients in $\sqrt{e}=a+b\sqrt{d}$ proves that two such fields coincide precisely when $e/d$ is a square. The only unramified member is ${\mathbb{Q}}_2(\sqrt{5})$: $(1+\sqrt{5})/2$ is integral with irreducible reduction $X^2+X+1$. Conversely any unramified quadratic extension has that residue field, and Hensel lemma lifts a residue root of $X^2-X-1$, recovering the same field.

In $L={\mathbb{Q}}_2(\sqrt{2})$, set $\pi =\sqrt{2}$ and normalize $v_L(\pi )=1$, so $v_L(2)=2$. The two terms of $a+b\pi$ have valuations of opposite parity, giving ${\mathcal{O}}_L={\mathbb{Z}}_2\oplus {\mathbb{Z}}_2\pi$. The nonidentity automorphism sends $\pi$ to $-\pi$, and $v_L(-2\pi )=3$. The lower ramification numbering therefore gives the displayed groups. The different exponent is three.